= Corrections to a pulling measurement of kinetic friction
{title2=$\mu_k=(F\cos\alpha-ma)/(mg-F\sin\alpha)$}
For a block sliding right on a stationary horizontal plane, take rightward <acceleration> $a$ as positive. A gauge reading $F$ for a pull at <angle> $\alpha$ above horizontal has horizontal component $F\cos\alpha$ and upward component $F\sin\alpha$. With no vertical <acceleration> or other <forces>, <Newton's second law> gives
$$
N=mg-F\sin\alpha,\qquad F\cos\alpha-|f_k|=ma,
$$
and therefore
$$
\mu_k=\frac{F\cos\alpha-ma}{mg-F\sin\alpha}.
$$
This inference requires continued contact with $N>0$ and rightward <contact slip>, so its numerator must be nonnegative. An upward pull reduces <normal force>; a downward pull, represented by $\alpha<0$, increases it. The horizontal constant-<velocity> formula is the special case $\alpha=0$, $a=0$.
For the original $2\,\mathrm{kg}$, $\mu_k=0.30$ model, an upward pull at $\alpha=30^\circ$ maintaining constant <velocity> needs
$$
F=\frac{\mu_kmg}{\cos\alpha+\mu_k\sin\alpha}\simeq5.7873\,\mathrm N,\qquad N\simeq16.7064\,\mathrm N.
$$
Dividing this gauge reading by $mg$ would give about $0.2953$, rather than $0.30$: the gauge reads the full inclined <force>, and the pull has changed the <normal force>.
Even with a horizontal pull, an accelerating measurement needs the $ma$ correction: $\mu_k=(F-ma)/(mg)$. In the same model, a horizontal $7.88\,\mathrm N$ pull produces $a=1\,\mathrm{m/s^2}$ while the block slides right; subtracting $ma=2\,\mathrm N$ recovers the $5.88\,\mathrm N$ <kinetic friction>. Dividing the uncorrected reading by $19.6\,\mathrm N$ instead gives about $0.4020$. Measure <acceleration> in an <inertial frame>, along the sliding direction, over the same interval as the gauge reading. These corrections follow from the <force> balance, independently of whether the resulting <coefficient of kinetic friction> is constant across runs.
// Target: classical-mechanics.bigb
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